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Ta có:
\(B=115\Rightarrow1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{x}\left(1+2+3+4+...+x\right)=115\)
\(\Leftrightarrow1+\dfrac{1}{2}.\dfrac{3.2}{2}+\dfrac{1}{3}.\dfrac{4.3}{2}+...+\dfrac{1}{x}.\dfrac{\left(1+x\right).x}{2}=115\)
\(\Leftrightarrow\dfrac{1}{2}.2+\dfrac{1}{2}.3+\dfrac{1}{2}.4+...+\dfrac{1}{2}.\left(x+1\right)=115\)
\(\Leftrightarrow\dfrac{1}{2}.\left(1+2+3+4+...+x\right)=115\)
\(\Leftrightarrow\dfrac{1}{2}.\dfrac{\left(x+1\right).x}{2}=115\Rightarrow\dfrac{x.\left(x+1\right)}{4}=115\)
\(\Rightarrow x.\left(x+1\right)=4.115=460\)
Đến đây thì phân tích 460 thành tích của 2 số tự nhiên liên tiếp nhưng ko đc.
a) Bổ sung cho đầy đủ đề
b) (3x - 1)/4 = (2x - 5)/3
3(3x - 1) = 4(2x - 5)
9x - 3 = 8x - 20
9x - 8x = -20 + 3
x = -17
c) Điều kiện: x ≠ -1/3
3/(-2) = (x - 3)/(3x + 1)
3.(3x + 1) = -2(x - 3)
9x + 3 = -2x + 6
9x + 2x = 6 - 3
11x = 3
x = 3/11 (nhận)
Vậy x = 3/11
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
b: =>x-1/2=1/3
=>x=5/6
c: =>2/3x-1=0 hoặc 3/4x+1/2=0
=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3
d =>4/9:x=10/3:9/4=10/3*4/9=40/27
=>x=4/9:40/27=4/9*27/40=108/360=3/10
Bài 1:
a.
$|x+\frac{7}{4}|=\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)
b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$
$|2x+1|=\frac{11}{15}$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)
c.
$3x(x+\frac{2}{3})=0$
\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)
d.
$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$
$\Leftrightarrow x=\frac{2}{5}$
Nguyễn Quý Trung:
\(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
Bạn bớt 2 vế đi 1/3 thì \(x=\dfrac{2}{5}\)
\(B=1+\dfrac{1}{2}\left(1+2\right)+...+\dfrac{1}{x}\left(1+2+..+x\right)\)
\(B_x=\dfrac{1}{x}\left(\dfrac{x\left(x+1\right)}{2}\right)=\dfrac{x+1}{2}\)
\(2B=2+3+4+5+...+\left(x+1\right)\)
\(2B+1=1+2+...+\left(x+1\right)=\dfrac{\left(x+1\right)\left(x+2\right)}{2}\)
\(B=115\Leftrightarrow2B+1=231\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=231.2=462\)=21.22
x=20